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Nursing Midterm Review Guide, Exams of Nursing

Atlantic Institute of Oriental Medicine (ATOM)Nursing

Nursing Midterm Review Guide Nursing Midterm Review Guide

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2025/2026

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Nursing Midterm Review Guide
Chapter 1-4
1. Which of the following is a condition which may occur during the incubation period?
-Transmission of infection.
2. Chicken pox is a highly communicable disease. It may be transmitted by direct contact with a
person infected with the varicella-zoster virus (VZV). The typical incubation time is between
10- 20 days. A boy started school 2 weeks after showing symptoms of chicken pox including
mild fever, skin rash, and fluid-filled blisters. One month after the boy returned to school, non of
his classmates had been infected by VZV. The main reason was:
-Contact was after infectious period.
3. The ability of a single person to remain free of clinical illness following exposure to
infectious agent is known as:
-immunity
4. Which of the following is characteristic of single-exposure, common-vehicle outbreak?
-The epidemic curve has a normal distribution when plotted against the logarithm of time.
Involve a sudden, rapid increase in cases of disease that are limited to persons who share
a common exposure.
5. What is the diarrhea attack rate in persons who ate both ice cream and pizza?
-39/52 (The attack rate is defined as the number of people who develop diarrhea divided by the
total number of people at risk.
6. What is the OVERALL attack rate in persons who did not eat ice cream?
-33% (The attack rate is the # of persons with diarrhea (14+9) divided by the total # of persons
who did not eat ice cream (40+30)
7. Which of the food items (or combination of items) is most likely to the infective item(s)?
-ice cream only (only 70% developed diarrhea regardless of their pizza consumption (39/52 and
11/15). Among both groups of persons who did not eat ice cream, each attack rate was equal to
or less than 35% (14/40 and 9/30).
8. Which of the following reasons can explain why a person who did not consume the
infective food item got sick?
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Nursing Midterm Review Guide

Chapter 1- 4

  1. Which of the following is a condition which may occur during the incubation period? -Transmission of infection.
  2. Chicken pox is a highly communicable disease. It may be transmitted by direct contact with a person infected with the varicella-zoster virus (VZV). The typical incubation time is between 10- 20 days. A boy started school 2 weeks after showing symptoms of chicken pox including mild fever, skin rash, and fluid-filled blisters. One month after the boy returned to school, non of his classmates had been infected by VZV. The main reason was: -Contact was after infectious period.
  3. The ability of a single person to remain free of clinical illness following exposure to infectious agent is known as: -immunity
  4. Which of the following is characteristic of single-exposure, common-vehicle outbreak? -The epidemic curve has a normal distribution when plotted against the logarithm of time. Involve a sudden, rapid increase in cases of disease that are limited to persons who share a common exposure.
  5. What is the diarrhea attack rate in persons who ate both ice cream and pizza? -39/52 (The attack rate is defined as the number of people who develop diarrhea divided by the total number of people at risk.
  6. What is the OVERALL attack rate in persons who did not eat ice cream? -33% (The attack rate is the # of persons with diarrhea (14+9) divided by the total # of persons who did not eat ice cream (40+30)
  7. Which of the food items (or combination of items) is most likely to the infective item(s)? -ice cream only (only 70% developed diarrhea regardless of their pizza consumption (39/52 and 11/15). Among both groups of persons who did not eat ice cream, each attack rate was equal to or less than 35% (14/40 and 9/30).
  8. Which of the following reasons can explain why a person who did not consume the infective food item got sick?

-All of the above (They were directly exposed to persons who did eat the infective food item, diarrhea is general symptom consistent with a number of illnesses, there may have been an inaccurate recall of which foods were eaten)

  1. An outbreak of gastroenteritis occurred at a boarding school with a student enrollment of 846. 57 students reported symptoms including vomiting, diarrhea, nausea, and low-grade fever between 10 p.m. on 9/24 and 8 p.m. on 9/25. The ill students lived in dormitories that housed 723 of the students. The table below provides information of the # of students per type of residence and the # reporting illnesses consistent with the described symptoms and onset time. Calculate the attack rate among all students at the boarding school. -6.7% (divide the total number of cases (57) by the total number of students (846).
  2. Same question above + Calculate the attack rates for boys and girls separately. 43 boys/426 total boys = 14 girls/420=
  3. Same question + What is the proportion of total cases occurring in boys? -43/57 = 75.4%
  4. Same question + What is the proportion of total cases occurring in students who live in dormitories? -52/57 = 91.2%
  5. Same question + Which proportion is more informative for the purpose of the outbreak investigation? -Both proportions are useful. Dormitory residents account for over 90% of the cases indicating an outbreak of an infectious agent that was transmitted at the school. Furthermore, over 75% o the cases were boys indicating that the responsible agent was more likely to have been transmitted in the boys dormitory.
  6. A group of researchers are interested in conducting a clinical trial to determine whether a new cholesterol-lowering agent was useful in preventing coronary heart disease (CHD). They identified 12,327 potential participants for the trial. At the initial clinical exam, 309 were discovered to have CHD. The remaining subjects entered the trial and were divided equally into the treatment and placebo groups. Of those in the treatment group, 505 developed CHD after 5 years of follow-up while 477 developed CHD during the same period in the placebo group. What was the prevalence of CHD at the initial exam? -The prevalence of CHD at the initial exam was 309 cases of CHD divided by 12, 327 participants. This equals a prevalence of 25.1 cases of CHD per 1,000 persons.
  1. Which of the following is an advantage of active surveillance? -Entails a concerted effort to collect information about disease occurrence. Involves dedicated staff members who have been specifically directed to contact physicians and hospitals in order to collect reports of disease cases.
  2. The population of a city on February 15, 2005, was 36,000. The city has a passive surveillance system that collects hospital and private physician reports of influenza cases every month. During the period between 1/1 and 4/1, 2005, 2,200 new cases of influenza occurred in the city. Of these cases, 775 persons were ill with influenza according to surveillance reports on 4/1. The monthly incidence rate of active cases o influenza for the 3-month period was:
  • 20 per 1,000 population (monthly incidence rate is calculated based on the # of new cases of a disease developing during the 3-month period.
  1. Same info + The prevalence rate of active influenzas as of 4/ 1/ 05.
  • 20 per 1,
  1. Same info –What can be inferred about influenza cases occurring in the city?
  • The average duration of influenza is approximately 1 month
  1. A study found that adults older than age 50 had a higher prevalence of pneumonia than those who were younger than age 50. Which of the following is consistent with this finding? -Incidence rates do not vary by age, but older adults have pneumonia for a longer duration compared to younger adults.
  2. Which of the following statements are true? More than one answer may be correct.
  • Prevalence rates are useful for public health planning -Incidence rates can be used to estimate prevalence when the mean duration of the disease is known.
  1. A disease has an incidence of 10 per 1,000 persons per year, and 80% of those affected will die within 1 year. Prior to the year 2000, only 50% of cases of the dease were detected by physician diagnosis prior to death. In the year 2000, a lab test was developed that identified 90% of cases an average of 6 months prior to symptom onset; however, the prognosis did not improve after diagnosis. Comparing the epidemiology of the disease prior to 2000 with the epidemiology of the disease after the development of the lab test, which statement is true concerning the disease in 2000? -Incidence is higher and prevalence is higher than in 1999.
  1. Same info- Which statement is true concerning the duration of the disease after the development of the lab test?
  • Mean duration of a case of the disease is longer in 2000
  1. Same info- Which statement is true concerning the disease-specific mortality rate after the development of the lab test? -The mortality rate for the disease is the same in 2000.
  2. In a coastal area of a country in which a tsunami struck, there were 100,000 deaths in a population of 2.4 million for the year ending 12/31/2005. What was the all-cause crude mortality rate per 1,000 persons during 2005? -41.7 per 1,000 persons. The rate is calculated by dividing 100,000 deaths by the population of 2,400,000 persons. To express as a rate per 1,000, the rate is multiplied by 1,000. 100,000/2,400,000 X 1,000=
  3. In an industrialized nation, there were 192 deaths due to lung disease in miners ages 20 to 64 years. The expected number of deaths in this occupational group, based on age-specific death rates for lung disease in all males ages 20 to 64 years, was 238 during 1990. What was the standardized mortality ratio (SMR) for lung diseases in miners? SMR= observed death for an occupation-cause-race group/expected deaths for an occupation- cause-race group X 100 192/238 X 100= 81%
  4. In 2001, a state enacted a law that required the use of safety seats for all children under 7 years of age and mandatory seatbelt use for all persons. The table below lists the number of deaths due to MVAs and the total population by age in 2000 (before the law) and in 2005 (4 years after the law was enacted). What is the age-specific mortality rate due to MVAs for children ages 0 to 18 years in 2000? -149/24,500 X1000= 6.1 per 1,
  5. Same info- using the pooled total of the 2000 and 2005 populations as the standard rate, calculate the age-adjusted mortality rate due to MVAs in 2005.
  • 2.3 MVA deaths per 1,000 persons. The key to calculating the age-adjusted rate is to pool the observed numbers for both time periods and to calculate the expected numbers of deaths in the 2005 population assuming that a common rate applied to the population. For those under 7 years, the pooled rate equals (44+20)/ (3,500 + 4,000). The pooled rate for this group is 8.5 per 1,000 persons. When this rate is multiplied by the 4,000 children under 7 years of age in 2005, the expected number of deaths is 34.13. Performing the same calculation for each
  • A mortality rate is an example of an incidence rate A mortality rate can approximate an incidence rate under conditions of a high case-fatality rate and a short duration of disease.
  1. Among those who are 25 years of age, those who have been driving less than 5 years had 13,700 motor vehicle accidents in 1 year, while those who had been driving for more than 5 years had 21,680 motor vehicle accidents during the same time period. It was concluded from these date that 25-year-olds with more driving experience have increased accidents compared to those who started driving later. This conclusion is:
  • Incorrect because rates are not reported. The information provided only enumerates motor vehicle accidents in two groups. In order to fully compare these counts, information is needed on the denominator, the number of persons driving in each group, so that rates can be calculated.
  1. For a disease such as liver cancer, which is highly fatal and of short duration, which of the following statements is true? Choose the best answer.
  • Incidence rate will be equal to mortality rates. Since the 5-year survival rate for liver cancer is 4%, most incident cases of liver cancer will result in a premature mortality. In this case, the mortality and incidence rates will be approximately equal.
  1. The prevalence rate of a disease is two times greater in women than in men, but the incidence rates are the same in men and women. Which of the following statements may explain this situation?
  • The case-fatality rate is lower for women Since men and women develop the disease at the same rate, the survival rate in women must be increase in order to increase duration and prevalence. A low case-fatality rate would contribute to an increased duration of the disease.
  1. The table below describes the number of illnesses and deaths caused by plague in four communities. The case-fatality rate associated with plague is lowest in which community?
  • Community C The case-fatality rate equals the number of deaths occurring from plague divided by all persons with the plague. In community c, the CFR is 300 divided by 400, or 60%. This lower than A (67%), B (75%), and D (77%).
  1. Same info- The proportionate mortality ratio associated with plague is lowest in which community?

-Community D, the proportionate mortality rate equals the number of deaths occurring from plague divided by all persons with the plague. In community D, the PMR if 500/5,000, or 10%. This is lower than A(50%), B(75%), and C(38%) Chapter 5- 6

  1. In a community-based hypertension testing program called HT-Aware, the detection level for high blood pressure is set at 140mmHg for systolic blood pressure. A separate testing program called HT-Warning in the same community sets the level at 130mmHg for high systolic blood pressure. Which statements are likely to be true?
  • The sensitivity of HT-Warning is greater than that of HT-Aware and the number of false positives is greater with HT-Warning than with HT-Aware
  1. A school nurse examined a population of 1,000 children in an attempt to detect nearsightedness. The prevalence of myopia in this population is known to be 15%. The sensitivity of the examination is 60% and its specificity is 80%. All children labeled as “positive” (i.e., suspected of having myopia) by the school nurse are sent for examination by an optometrist. The sensitivity of the optometrist’s examination is 98% and its specificity is 90%. How many children are labeled “positive” by the school nurse? -There are 150 children with myopia in the school population (15% prevalence among 1, children). The school nurse will identify 60% of those who truly have the condition, or 90 cases (60% sensitivity multiplied by 150 myopic children). Further, the school nurse will incorrectly identify 170 false positive cases of myopia among those who do not have the condition (80% specificity multiplied by 850 non-myopic children). The sum of the cases labeled as positive by the school nurse equals 260 children (90 true myopic children plus 170 false positive children).
  2. Same info- What is the positive predictive value (PPV) of the school nurses’s exam?
  • PPV is 34.6% (90 true myopic children /260 children labeled as myopic by the school nurse. The PPV of the school nurse’s exam is equal to the number of true positive cases divided by the total number of those that the school nurse labels as positive.
  1. How many children will be labeled myopic following the optometrist’s exam? -Since the optometrist will only test children who have been labeled as myopic by the school nurse, the testing group for this sequential exam is 260 children. The optometrist labels 105 children as myopic. Among the 90 myopic children correctly referred by the school nurse, the optometrist identifies 88 of them as myopic (98% sensitivity multiplied by 90 true cases of myopia). Further, the optometrist will incorrectly identify 17 false positive cases among the 170 children referred by the school nurse who do not have myopia. The sum of the cases labeled as positive by the optometrist equals 105 children (89 true cases plus 17 false positive cases).
  2. What is the positive predictive value (PPV) of the optometrist’s exam?

each group is shown in the table. What is the sensitivity of the PSA screening test in the combined groups? -The sensitivity equals the number of true positives detected among all true positives. Since a biopsy is the gold standard test for prostate cancer, all 1,500 men in group A are positive for prostate cancer. The PSA test indicated that 1,155 of these men had prostate cancer, a sensitivity of 77%.

  1. Same info- what is the specificity of the screening test in the combined groups? The specificity equals the number of true negatives detected among all true negatives. Among the 3,000 men who did not have prostate cancer, the test correctly identified 2,760 men as negative for prostate cancer (3,000 minus 240 false positives). This gives a sensitivity of 92%.
  2. What is the positive predictive value (PPV) of the screening test in the combined groups? The PPV is 83%. This value is found by dividing 1,155 true positives by the total number of all positives indicated by the PSA test (1,155 plus 240).
  3. The PSA screening test is used in the same way in two equal-sized populations of men living in different areas of the United States, but the proportion of false positives among those who have a positive PSA test in the first population is lower than that among those who have a positive PSA test in the second population. What is the likely explanation for this finding?
  • The prevalence of disease is higher in the first population We can assume that the specificity of the test will be similar in each population. Therefore the proportion of false positives found among the true negatives should be the same in each population. However, the proportion of false positives among all positives on the PSA screening test will be influenced by the number of true positives detected by the test. Since the sensitivity of the test will also be the same, we can assume that more true positives exist in the population of men with a lower proportion of false positive tests due to an increase in the PPV.
  1. Test A has a sensitivity of 95% and a specificity of 90%. Test B has a sensitivity of 80% and a specificity of 98%. In a community of 10,000 people with 5% prevalence of the disease, Test A has always been given before Test B. What is the best reason for changing the order of the tests?
  • The total number of false positives found by both tests is decreased if Test B is given first. A sequential testing process would only refer those with positive results to the second test. Since Test B has a higher specificity, then fewer false positives will be referred for Test A, thereby decreasing the number of false positives found. This can be shown by calculation if we assume that 500 persons have the disease among the 10,000 in the population. Test B will find only 190 false positives for referral (9,500 true negatives less the number of true negatives multiplied by 98% specificity). Performing test A first results in 950 false positives referred for the second test (9,500 true negative less the number of true negatives multiplied by 90% specificity).
  1. Two neurologists, Drs. J and K, independently examined 70 magnetic resonance images (MRIs) for evidence of brain tumors. As shown in the table below, the neurologists read each

MRI as either “positive” or “negative” for brain tumors. Based on the above information, the overall percent agreement between the two doctors including all observations is: -62.9% The two doctors agree on 44 of the 70 MRI readings. This includes the 26 that they both labeled as positive for brain tumors and the 18 that they both agreed were negative for brain tumors.

  1. What is the estimate of kappa for the reliability of the two doctors’ test results?
  • 24.9% The estimate of kappa expresses the observed agreement of two testers in excess of chance alone. It is found by applying the expected agreement rates for both testers. In this case, Dr. K labeled 38 of the 70 MRIs as positive (54.3% of all MRIs) and 32 as negative (45.7% of all slides). Dr. J labeled 57.1% of the MRIs as positive (40 of 70) and 42.9% as negative. We would expect that if Dr. K had the same rate of positive and negative findings as Dr. J then they would agree by chance on 21.7 of the 38 positive MRIs that were found (38 multiplied by 0.571). Further, they would agree by chance on 13.7 of the 32 negative MRIs that were found ( multiplied by 0.429). Therefore, we would expect the two doctors to agree by chance on 50.6% of the MRIs (21.7 positive agreements plus 13.7 negative agreements equals 35.4, then divide this by the total of 70 to get an expected overall agreement of 50.6%). Now, kappa can be calculated as the observed agreement less expected divided by 100% less the expected agreement — in this instance, 62.9% minus 50.6% divided by 100% less 50.6%. 12.3% divided by 49.4% results in a kappa of 24.9%.
  1. This table represents the results of coronary magnetic resonance (CMR) angiography compared to x-ray angiography (the gold standard in diagnosis of coronary artery disease) in a high-risk population of patients scheduled to undergo x-ray angiography for suspected coronary artery disease. In the general population, the prevalence of coronary artery disease is approximately 6%. Assuming that this sample of patients is representative of the general population, the sensitivity of the CMR test in the general population would be approximately:
  • Between 90% and 95% If we assume that the prevalence of disease is similar, then we can accept that 60 persons with a positive x-ray will be true cases of coronary artery disease. In this instance, the CMR test positively identifies 56 of the 60 true cases, a sensitivity of 93.3%.
  1. After reviewing the results of the test comparison, an epidemiologist decides that the specificity of the test is too low. Using the same CMR images, he raises the cutoff value for a positive test to increase the specificity. What is the likely effect on the sensitivity? -Sensitivity will decrease The increase in the cutoff value for a positive test will reduce the sensitivity of the test even though the specificity is increased. This will result in the misidentification of true positive cases as false negatives if their CMR values are below the cutoff value suggested by the epidemiologist.

example, we have 950 persons alive at the beginning of year 2 (thus, the end of year 1). Of this group, 30 die by the end of the second year. This gives a survival rate of 920 divided by 950, or 97%.

  1. What was the cumulative probability of surviving after only 2 years of follow-up? -The cumulative survival is the total number of those surviving by the end of the second year divided by all persons who were alive at the beginning of follow-up. In this example, there were 920 survivors among the 1,000 persons who were alive at the beginning of observation. This equals a cumulative survival of 92%. Alternatively, this cumulative survival can be calculated by multiplying the survival rates for each period of interest. In this example 95% survival for year 1 multiplied by 97% survival for year 2 equals a cumulative survival of 92%.
  2. An important assumption in this type of analysis is that: -No change has occurred in the effectiveness of treatment during the 3-year period. An important assumption of survival analysis is that separate strata, in this example, years of follow-up, have similar underlying rates of survival. If some external factor were to differentially influence survival during a portion of the follow-up time then we would not be able to assume a cumulative survival that is consistent during the entire study period.
  3. /30. Tables not included
  4. Which of the following is a key assumption involved in the use of life-table analysis? -The risk of disease does not change within each interval over the period of observation Life-table analysis depends upon a consistent rate of survival during all periods of the study. Changes in the rate of survival may be due to external influences that are operating at later times on only a portion of the initial population. Since those who have died earlier in the study period will not experience these external influences, the comparison between periods is rendered invalid.
  5. Which of the following is a measure of disease prognosis?
  • Median survival time Disease prognosis indicates the likelihood of survival once a disease has become manifest. The median survival time reflects the length of time that the 50th percentile of affected persons has. It differs from the mean survival time in that the mean survival time is an average that may be influenced by extremely low or high survival times. The median survival time consists of an ordering of all survival times with the midpoint of the distribution taken as the duration of survival.
  1. In 2003, Sudden Acute Respiratory Syndrome (SARS) appeared in several countries, mainly in Asia. The disease was determined to have been caused by a virus that could be spread from person –to person from the index case occurring in mainland China. This table reflects the total number of reported cases of SARS and deaths among those cases as best as can be determined. What is the overall case-fatality rate for the worldwide epidemic of SARS? -9.5% This can be found by dividing the total number of deaths due to SARS by the total number of cases. This equals a case-fatality rate of 9.5%.
  1. Based on the table, we can conclude that the case-fatality rate (CFR) in Vietnam: -Is almost one half that of the case-fatality rate in Singapore The CFR in Vietnam equals 5 divided by 63, or 7.9%, while that of Singapore equals 15%. This is approximately one half the rate.
  2. Following a revision in the case definition, more persons were found to have suffered from an infection with the SARS virus. The inclusion of these cases, almost all asymptomatic, did not impact the total number of SARS fatalities. What happened to the case-fatality rate (CFR) following this reclassification?
  • It was decreased The increase in prevalent cases with no change in mortality would decrease the CFR since the numerator, number of deaths due to SARS, would stay the same while the denominator, number of cases, increased.
  1. What is the probability of surviving the second year of the study given that a person survived the first year? The independent probability of surviving the second year for all persons who survived the first year is found by dividing the number of survivors at the end of the period by the total number present at the beginning of the period.
  2. table not avb
  3. What is the probability chance of surviving 3 years after diagnosis? The cumulative survival probability for all 3 years equals the product of the independent interval survival probabilities. In this example, 59% multiplied by 53% multiplied by 70.4% gives a cumulative survival probability of 22%.
  4. table not avb
  5. Before reporting the results of this survival analysis, the investigators compared baseline characteristics of the 44 people who withdrew from the study before its end to those who had complete follow-up. This was done: -To check whether those withdrawing from the study are similar to persons remaining in the study A key assumption in life table analysis is to insure that the experience of those lost to follow-up, or withdrawals from the study, is the same as those remaining under observation. Chapter 7- 8
  6. Which of the following statements best describe efficacy?
  • It is an estimate of the benefit of treatment under ideal conditions, and it is an estimate of the reduction of disease in treated groups.
  1. A randomized, double-blind clinical trial of a varicella vaccine observed an estimated incidence of 25% chickenpox episodes in persons receiving the vaccine, compared to 80% among persons receiving a placebo. The estimated efficacy of the vaccine is: -68.8% The efficacy is calculated by subtracting the rate in those vaccinated from the placebo rate and dividing by the placebo rate. In this example, 80% minus 25% equals 55%. Dividing by 80% results in an efficacy of 68.8%.
  2. A multicenter double-blind randomized study was carried out to compare the effect of drug X with that of a placebo in patients surviving acute myocardial infarction (AMI). Treatment with the drug started 7 days after infarction in 1,884 patients, 52% of all persons who were evaluated for entry into the study. 945 participants were randomized to treatment with drug X while 939 were assigned to the placebo group. Patients were then followed for 12 months for reinfarction. There were 152 deaths in the placebo group and 98 in the group receiving drug X. After entry into the study, patients were first classified into three groups, those who had a previous AMI, those with a first AMI who were at high risk for other cardiovascular diseases such as congestive heart failure, and those with a first AMI who were at low risk for other cardiovascular diseases. Which term best describes the study design?
  • Randomized clinical trial with stratified randomization
  1. After assignment to treatment group, 77% of those in the placebo group were men, while 80% of those in the drug X group were men. Which statement is most likely to be true? -Randomization was successful since the investigators did not alter the selection of participants in either group in order to ensure equal percentages of men Randomization is the process of assigning participants to a study group with nonpredictable assignment. Randomization does not guarantee equally comparable groups; however, it ensures that the researchers do not influence the selection of individual participants into a treatment group. Over large samples, characteristics will tend to be approximately equal, but this is not guaranteed since random chance can influence assignments.
  2. A preliminary analysis was conducted after 6 months and found that 87% of participants in the placebo group and 85% of those in the drug X group had taken more than 90% of their prescribed dosages. Which statement best describes this finding? -The characteristics of patients who failed to comply with the treatment dosages should be assessed as they may differ from those who complied. The compliance rates are high and are approximately equal between the two groups so we can expect the estimated effects of treatment to be reasonably valid. An analysis should be conducted of characteristics of non-compliers as they may differ in terms of important predictors of recurrent AMI from those who take the prescribed dosages.
  3. Which of the following statements best describes the reason for conducting the study as a double-blind trial?

-double blinding ensures that potential biases regarding selection, follow-up, and analysis can be reduced. Both participants and researchers can have preconceived biases regarding the potential effect of a studied treatment. By masking the assignment of subjects from themselves and the researchers, the effect that these biases may have on the comparative analyses can be reduced.

  1. The researchers conclude that treatment with drug X reduces mortality in patients who have had an AMI. The researchers are:
  • Correct because the rate of death is decreased in the drugX group The rate of death in the drug X group was 10.4% while it was 16.2% in the placebo group. Although a statistical test such as a chi-square test should be done, the large difference in these rates between two approximately equivalent groups indicates that drug X contributed to increased survivability compared to those receiving the placebo.
  1. The following data come from a study of approaches to smoking cessation. Smokers who want to quit were randomized to one of four groups: control group C who received no intervention assistance, quitting guide group Q who received brochures about how to quit smoking, quitting guide and support group QS who received quitting brochures as well as social support brochures listing benefits of smoking cessation, and telephone support group T who received the brochures and a monthly phone call from a counselor. Participants received mailed surveys at 8, 16, and 24 months after randomization. The results after 2 years are in the table below. What is the overall quit rate after 2 years of follow-up? -no table provided
  • The overall quit rate after 2 years is 17.6%. This is estimated by summing the total number of those who quit smoking (331) and dividing by all participants who returned the survey after 2 years (1877).
  1. same info- Which group had the least success in terms of quitting smoking? -Group QS. Each group has approximately equal numbers of participants with missing information. Therefore, the rates for quitting by group are 18.1% for group C, 15.2% for group Q, 14.2% for group QS, and 23% for group T. Group QS had the lowest rate of quitting among the four groups.
  2. same info- What is the main purpose of randomization in this study? -To avoid assigning more persons who have tried and failed to quit in the past to the control group
  • The goal of randomization is to reduce selection bias in the assignment of study groups. Obviously, a participant’s previous history with smoking cessation is an important predictor of his future success. In order to avoid assigning more persons who had tried and failed in the past to one specific group, randomization of all participants is the best approach.

be compromised by biases regarding the investigator’s theorized association of the exposure and disease.

  1. In a study of the adverse effects of x-rays among children, a retrospective cohort study was done using records from several large children’s hospitals for the period of 1980 to 1985. 10, children were selected as a representative population of ill children seen at the hospitals during that time. Subjects were classified according to whether or not they received an x-ray during their stay in the hospital and were followed from their hospital stay through 2005 for the development of cancer. During the follow-up period, 49 incident cancers occurred in 3, children who had received an x-ray during their hospitalization. In this retrospective study, which of the following groups are eligible for selection into the study?
  • Children receiving x-rays for broken bones in 1983. Children who received an x-ray during the defined study period and did not have a cancer diagnosis during that time can be included in the study.
  1. In a study of the adverse effects of x-rays among children, a retrospective cohort study was done using records from several large children’s hospitals for the period of 1980 to 1985. 10,000 children were selected as a representative population of ill children seen at the hospitals during that time. Subjects were classified according to whether or not they received an x-ray during their stay in the hospital and were followed from their hospital stay through 2005 for the development of cancer. During the follow-up period, 49 incident cancers occurred in 3, children who had received an x-ray, and 47 incident cancers occurred in the 6,737 children who had not received an x-ray during their hospitalization. What are the rates of cancer incidence in each exposure group?
  • The rate of cancer incidence in the x-ray exposed group is 15 per 1,000 (49 divided by 3, multiplied by 1,000) and 7 per 1,000 in the nonexposed group (47 divided by 6,737 multiplied by 1,000). 15 per 1000 in xray exposed 7 per 1000 in nonexposed xray
  1. Same info as above. What is the attributable risk of cancer due to x-ray in this study population? What is the interpretation of this estimate?
  • 15-7=8 out of the 15 xray exposed The attributable risk equals the incidence rate in the exposed group minus the incidence rate in the nonexposed group. In this instance, the attributable risk is 8 per 1,000. This estimate is interpreted to mean that 8 of the 15 incident cases of cancer occurring in 1,000 children exposed to x-rays are due to the exposure itself.
  1. Same info as above. What is the risk ratio for the effect of exposure on the development of cancer in this study? What is the interpretation of this estimated ratio? -The risk ratio is found by dividing the rate of cancers for each exposure group. In this instance, 15 per 1,000 (0.015) divided by 7 per 1,000 (0.007) equals a risk ratio of 2.1. This indicates that the risk of cancer is twice as high in children who received x-rays during their stay in the hospital. .015/.007=2.
  1. Same info as above. Which of the following issues should the investigators consider when interpreting whether a causal association exists between cancer incidence and childhood x-ray? -ALL OF THE ABOVE -Some study subjects were treated for cancer starting in 1980 -Some study subjects had parents who were diagnosed with cancer -Some children received xrays at other hospitals not included in this study -The children were different ages when they were admitted to the hospital Each of these issues may be related to the exposure and cancer association that the investigators are reporting and thus they must be considered. First, subjects treated for cancers in their initial hospitalization should have been excluded from the study selection. Second, genetic predisposition to cancer is an important factor in cancer incidence. Third, misclassified exposure may influence the relationship, especially if children received x-rays at other hospitals but did not have this fact noted in their medical records at the study hospital. Finally, age of the children is an important factor related to x-ray dosage, development, and subsequent cancer incidence. All of these issues may lead to important differences between the two defined exposure groups.
  2. Which of the following may be a factor that would result from the inability to use randomization in a cohort study?
  • The possibility that a factor which leads to exposure may be causally associated with the disease. Randomization removes potential bias from the designation of treatment, or exposure, groups. Without this step, it may be likely that the exposed and nonexposed groups in a cohort study are misidentified due to other factors that are associated with exposure and subsequent disease outcomes.
  1. 6,750 people who were free of disease X were enrolled in a cohort study in 1985 and followed with annual exams and interviews through 1995. Exposure to factor A was determined at study enrollment and the participants were followed until 1995 to observe new cases of disease X. Data from the study at the end of follow-up are shown in the following table. What is the incidence rate of disease X among persons exposed to factor A?

The incidence rate for disease X among exposed persons is found by dividing 120 (the number of people with disease X) by the total number of exposed persons (120 plus 2880). If multiplied by 100 then the incidence rate would equal 4 per 100 persons; however, this metric is not explicitly requested.

  1. What is the relative risk for the effect of exposure to factor A on disease X?

The relative risk equals the incidence rate in the exposed divided by the incidence rate in the not exposed. In this example, the incidence rate in the exposed group is 0.04 divided by the incidence rate in the not exposed (0.008). The ratio is equal to 5.0 in this example. This indicates that the incidence rate of disease X is 5 times higher in those exposed to factor A.

  1. In 2002, investigators started a study of the association of cholesterol levels and stroke in a group of 2,000 healthy persons who had participated in a cholesterol screening program in